LeetCode - 139. Word Break

Medium

題目

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

Example 1

Input: s = "leetcode", wordDict = ["leet","code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2

Input: s = "applepenapple", wordDict = ["apple","pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
Note that you are allowed to reuse a dictionary word.

Example 3

Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
Output: false

Constraints

1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
s and wordDict[i] consist of only lowercase English letters.
All the strings of wordDict are unique.

解題思路

經典字串 DP 題
判斷 s 能不能被切成多個存在於 wordDict 裡的字串

DP 定義：
- dp[i]：代表 s[0...i-1] 是否可以被成功切分
初始化：
- dp[0] = true
- 代表空字串可以被成功切分
先將 wordDict 存進 unordered_map
- 方便後面快速查詢某段 substring 是否存在於字典中
狀態轉移：
- 枚舉結尾位置 i
- 再枚舉切割點 j
- 如果：
  - dp[j] == true
  - 代表 s[0...j-1] 可以被成功切分
  - 且 s.substr(j, i-j) 存在於 wordDict
  - 代表 s[j...i-1] 也是合法字串
- 那麼：
  - dp[i] = true
也就是說：
- 前半段可以被成功切分
- 後半段又是一個字典裡的字
- 所以目前這段也可以被成功切分
最後答案為：
dp[n]

Solution Code

Time Complexity
O(n^3)

Space Complexity
O(n + wordDict.length)

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        int n = s.size();
        vector<bool> dp(n+1, false);

        unordered_map<string, bool> mp;
        for(auto word: wordDict) mp[word] = true;

        dp[0] = true;
        
        for(int i = 1; i <= n; ++i){
            for(int j = 0; j < i; ++j){
                // dp[j] = true 代表: s[0...j-1] 可以被成功切分
                // mp[s.substr(j, i-j)] 代表: s[j...i-1] 是不是 wordDict 裡的字串
                if(dp[j] && mp[s.substr(j, i-j)]) dp[i] = true;
            }
        }
        return dp[n];
    }
};